A dice is rolled twice . Find the probability that
1 . 5 will not come up either time.
2. 5 will come up exactly one time.
Answers
ANSWER
In a throw of pair of dice, total no of possible outcomes=36(6×6) which are
(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)
(2,1)(2,2)(2,3)(2,4)(2,5)(2,6)
(3,1)(3,2)(3,3)(3,4)(3,5)(3,6)
(4,1)(4,2)(4,3)(4,4)(4,5)(4,6)
(5,1)(5,2)(5,3)(5,4)(5,5)(5,6)
(6,1)(6,2)(6,3)(6,4)(6,5)(6,6)
Solution(i):
Let E be event of getting 5 on either time
No. of favorable outcomes =11 (i.e.,(1,5)(2,5)(3,5)(4,5)(5,1)(5,2)(5,3)(5,4)(5,5)(5,6)(6,5))
We know that, Probability P(E) =
(Total no.of possible outcomes)
(No.of favorable outcomes)
=
36
11
We have, P(E)+P(
E
ˉ
)=1
P(
E
ˉ
)=1−P(E)=1−
36
11
=
36
25
Therefore, the probability of not getting 5 either time =
36
25
Solution(i):
Let E be event of getting 5 exactly one time
No. of favorable outcomes=10 (i.e.,(1,5)(2,5)(3,5)(4,5)(5,1)(5,2)(5,3)(5,4)(5,6)(6,5))
We know that, Probability P(E) =
(Total no.of possible outcomes)
(No.of favorable outcomes)
=
36
10
=
18
5
Therefore, the probability of getting 5 exactly one time =
18
5
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